Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a question about joint convergence results derived from an FCLT (i.e., a Functional Central Limit Theorem). To motivate my question, consider the following setup:

Let $y_t$ be a random walk $$y_t = \rho y_t + \epsilon_t ,\quad \rho = 1$$ where $(\epsilon_t)$ satisfies a FCLT, i.e., $$ \frac{1}{\sqrt T} \sum_{t=1}^{[rT]}\epsilon_t\Longrightarrow \sigma W(r),$$ where $W$ denotes a standard Brownian motion on $[0,1]$, $\sigma>0$ is a constant, and $\Longrightarrow$ denotes weak convergence. Then I can follow standard textbook arguments (using the FCLT and the Continuous Mapping Theorem) to derive the following convergence results:

  • $$T^{-1} \sum_{t=1}^T y_{t-1}\epsilon_t\Longrightarrow \sigma^2\int_0^1W(t)dW(t)$$
  • $$T^{-3/2} \sum_{t=1}^T y_{t-1}\Longrightarrow \sigma\int_0^1W(t)dt$$
  • $$T^{-2} \sum_{t=1}^T y_{t-1}^2\Longrightarrow \sigma^2\int_0^1W(t)^2dt$$

So no problem so far. But very often (in research papers and textbooks), you see the following result:

$$\left(\begin{array}{c} T^{-1} \sum_{t=1}^T y_{t-1}\epsilon_t \\ T^{-3/2} \sum_{t=1}^T y_{t-1} \\ T^{-2} \sum_{t=1}^T y_{t-1}^2 \end{array}\right) \Longrightarrow \left(\begin{array}{c} \sigma^2\int_0^1W(t)dW(t) \\ \sigma\int_0^1W(t)dt \\ \sigma^2\int_0^1W(t)^2dt \end{array} \right)$$

And this I do not understand (i.e., here is my question): HOW DO I DERIVE THE JOINT CONVERGENCE, provided I have the convergence of the marginals. I do know that convergence of the marginals in distribution (or weakly) is not sufficient for convergence of the vector in distribution (or weakly). Am I missing something? What tools, theorems, insights can I use to derive the above joint convergence? Many thanks for any help, I really appreciate everything you can give me!

Cheers!

share|improve this question

1 Answer 1

For each positive $T$, consider the process $x_T=(x_T(s))_{0\leqslant s\leqslant 1}$ defined by $$ x_T(s)=\frac{y_{\lfloor sT\rfloor}}{\sigma\sqrt{T}}, $$ for every $0\leqslant s\leqslant 1$. The hypothesis is that $x_T$ converges in distribution to a standard Brownian motion $W=(W(s))_{0\leqslant s\leqslant 1}$ when $T\to\infty$. Hence, for every suitable functional $G$, $G(x_T)$ converges in distribution to $G(W)$.

Our goal is to use this for some specific three-dimensional functional $G$. The two last components of $G$ are obvious but the first one requires some more massaging. To this end, note that $$ \frac2T\sum_{t=1}^Ty_{t-1}\epsilon_t=\frac1Ty_T^2-\eta_T=\sigma^2x_T(1)^2-\eta_T,\qquad\eta_T=\frac1T\sum_{t=1}^T\epsilon_t^2. $$ Assume that $\eta_T\to\sigma^2$ almost surely (this holds, for example, if $(\epsilon_t)_t$ is i.i.d. centered with variance $\sigma^2$). Then the case which interests you is $$ G(x)=\left(\frac{x(1)^2-1}2,\int_0^1x(s)\mathrm ds,\int_0^1x(s)^2\mathrm ds\right). $$ Indeed, note for example that $$ \int_0^1x_T(s)^2\mathrm ds=\frac1T\frac1T\sum_{t=0}^{T-1}y_t^2. $$ This proves the convergence in distribution of the three-dimensional vector considered since, finally, $$ \frac{W(1)^2-1}2=\int_0^1W(s)\mathrm dW(s). $$ Nota: There exists simple counterexamples to the hypothesis that $\eta_T\to\sigma^2$. For example, if $\epsilon_t=a_t+(-1)^t$ where $(a_t)_t$ is i.i.d. and the distribution of $a_t$ is symmetric, then the FCLT holds with $\sigma^2=\mathbb E(a_1^2)$ but $\eta_T\to1+\mathbb E(a_1^2)\ne\sigma^2$.

share|improve this answer
    
many thanks for your very well explained answer. just one question (regarding the appearing $x(1)$ term in the first coordinate of $G$): are coordinate mappings in the Skorohod space continuous? –  s_2 Jan 15 '13 at 15:33

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.